将元组列表划分为多个列表的更简洁方法

Question

赏金将在 7 天后到期。 此问题的答案有资格获得+500声望赏金。 richardec想要奖励现有的答案。

标题是不言自明的，是否有一行表达式可以做到这一点？

可重现的例子：

Data = [(21,1,2000), (30,0,1991), (27,0,1994)]

age = [el[0] for el in Data]
sex = [el[1] for el in Data]
birth = [el[2] for el in Data]

print(age, sex, birth)

Answer 1

使用zip ：

age, sex, birth = zip(*Data)  # tuple

# OR

age, sex, birth = map(list, zip(*Data))  # list

>>> age
[21, 30, 27]

>>> sex
[1, 0, 0]

>> birth
[2000, 1991, 1994]

Answer 2

对@Corralien 解决方案稍作修改，要在list而不是在tuple获得结果，请使用map 、 zip一起获得结果

Data = [(21,1,2000), (30,0,1991), (27,0,1994)]
age, sex, birth = map(list, zip(*Data))
print(age, sex, birth, sep ='\n')

输出

[21, 30, 27] # age
[1, 0, 0] # sex
[2000, 1991, 1994] # birth

Answer 3

# You can do this by combining zip() with map() :

Data = [(21,1,2000), (30,0,1991), (27,0,1994)]
result = list(map(list, zip(*Data)))
print(result) 
# [[21, 30, 27], [1, 0, 0], [2000, 1991, 1994]]